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Hello,

I've been wondering today around the following exercise in J.Heinonen's "Lectures on Analysis on Metric Spaces" (see below for terminology): prove that the statement of Vitali Covering Theorem for bounded subsets $A\subset X$ implies the statement for all subsets $A\subset X$.

Here $X$ is a metric space equipped with a Borel regular outer measure $\mu$ which is doubling (balls have finite measure and doubling the radius of a ball increases its measure at most by a constant factor).

Let me quickly recall the actual statement of the theorem: in a doubling metric space $(X,\mu)$ with $\mu$ Borel regular, any family $\mathcal{F}$ of closed balls centerd at a set $A\subset X$ with the property that about every $a\in A$ there are elements in $\mathcal{F}$ of arbitrarily small radii admits a countable disjointed subfamily $\mathcal{G}$ covering almost all of $A$.

The actual proof of Heinonen works for sets of finite measure. The conclusion for unbounded sets is trivial for Lebesgue measure in $\mathbb{R}^n$, and whenever boundaries of balls have measure zero (so we can a.e partition $A$ by subsets admiting the conclusion of Vitali by mutually disjoint families). If boundaries of balls are not negligible (at least for some family of concentric balls with radii going to infinity) I can't find the argument bringing me there for arbitrary subsets.

I have seen a couple of references: Wikipedia (ok, not the best one maybe) works with Lebesgue measure, and Evans "Measure Theory and Fine Properties of Functions" does it for Radon Measures on $\mathbb{R}^n$, but only on sets of finite measure.

So before proceeding with trying the problem. Do you know whether this is true or have a hint for proving it?

Best

1

# Vitali Covering Theorem for Arbitrary Subsets of Doubling Metric Spaces

Hello,

I've been wondering today around the following exercise in J.Heinonen's "Lectures on Analysis on Metric Spaces" (see below for terminology): prove that the statement of Vitali Covering Theorem for bounded subsets $A\subset X$ implies the statement for all subsets $A\subset X$.

Here $X$ is a metric space equipped with a Borel regular measure $\mu$ which is doubling (balls have finite measure and doubling the radius of a ball increases its measure at most by a constant factor).

Let me quickly recall the actual statement of the theorem: in a doubling metric space $(X,\mu)$ with $\mu$ Borel regular, any family $\mathcal{F}$ of closed balls centerd at a set $A\subset X$ with the property that about every $a\in A$ there are elements in $\mathcal{F}$ of arbitrarily small radii admits a countable disjointed subfamily $\mathcal{G}$ covering almost all of $A$.

The actual proof of Heinonen works for sets of finite measure. The conclusion for unbounded sets is trivial for Lebesgue measure in $\mathbb{R}^n$, and whenever boundaries of balls have measure zero (so we can a.e partition $A$ by subsets admiting the conclusion of Vitali by mutually disjoint families). If boundaries of balls are not negligible (at least for some family of concentric balls with radii going to infinity) I can't find the argument bringing me there for arbitrary subsets.

I have seen a couple of references: Wikipedia (ok, not the best one maybe) works with Lebesgue measure, and Evans "Measure Theory and Fine Properties of Functions" does it for Radon Measures on $\mathbb{R}^n$, but only on sets of finite measure.

So before proceeding with trying the problem. Do you know whether this is true or have a hint for proving it?

Best